The Monotonic Sequence Theorem for Convergence

# The Monotonic Sequence Theorem for Convergence

We will now look at a very important theorem regarding bounded monotonic sequences (ones that are bounded above by $M$ or below by $m$ and are increasing or decreasing) and convergence of these sequences.

Theorem: If $\{ a_n \}_{n=1}^{\infty}$ is a bounded above or bounded below and is monotonic, then $\{ a_n \}_{n=1}^{\infty}$ is also a convergent sequence. |

**Proof of Theorem:**First assume that $\{ a_n \}_{n=1}^{\infty}$ is an increasing sequence, that is $a_{n+1} > a_{n}$ for all $n \in N$, and suppose that this sequence is also bounded, i.e., the set $A = \{ a_n : n \in \mathbb{N} \}$ is bounded above. We note that this sequence cannot be bounded below otherwise it would not be an increasing sequence. Suppose that we denote this upper bound $L$, and denote $L - \epsilon$ where $\epsilon > 0$ to be very close to this upper bound $L$. Therefore, for some $N \in \mathbb{N}$, it follows that for any $n > N$ since $\{ a_n \}$ is increasing, $a_n > L - \epsilon$.

- Therefore $a_n > L - \epsilon$ or rather $L - a_n < \epsilon$. But $\epsilon > 0$ so instead we get that $0 < L - a_n < \epsilon$. We thus obtain that $a_n < L$ when $n > N$, $\mid L - a_n \mid < \epsilon$ which implies that $\lim_{n \to \infty} a_n = L$.

- Now suppose that $\{ a_n \}_{n=1}^{\infty}$ is a decreasing sequence, that is $a_{n+1} < a_n$ for all $n \in N$, and suppose that this sequence is also bounded, i.e., the set $A = \{ a_n : n \in \mathbb{N} \}$ is bounded below. We note that this sequence cannot be bounded above otherwise it would not be a decreasing sequence. Suppose we denote this lower bound $L$ and denote $L + \epsilon$ where $\epsilon > 0$ to be very close to this upper bound $L$. Therefore for some $N \in \mathbb{N}$, it follows that for any $n > N$ since $\{ a_n \}$ is decreasing, $a_n < L + \epsilon$.

- Therefore $a_n < L + \epsilon$ or rather $a_n - L < \epsilon$. But $\epsilon > 0$ so instead we get that $0 < a_n - L < \epsilon$. We thus obtain that $a_n > L$ when $n < N$, $\mid a_n - L \mid < \epsilon$ which implies that $\lim_{n \to \infty} a_n = L$. $\blacksquare$